Question

Suppose the functions $ f(x) $ and $ g(x) $ are defined in this way:
 $
  f(x) = \beta {x^2} + 3,\; - \infty < x < - 1 \\
  f(x) = 2x + \alpha ,\; - 1 \leqslant x < \infty \\
  $ and $
  g(x) = x + 4,\;0 \leqslant x \leqslant 8 \\
  g(x) = - 3x - 2,\; - \infty < x < 0 \\
  $
If it is given that $ \alpha = 2,\;\beta = 3 $ , then identify the condition where the composition function: $ g(f(x)) $ is not defined:
(i) $ \alpha \in (10,\infty ),\;\beta \in (5,\infty ) $
(ii) $ \alpha \in (4,10),\;\beta \in (5,\infty ) $
(iii) $ \alpha \in (10,\infty ),\;\beta \in (0,1) $
(iv) $ \alpha \in (4,10),\;\beta \in (1,5) $

Answer 1

Hint: Carefully observe the functions given to us, and the interval of $x$in each definition. Remember that we are checking for the interval or points where the composition function $g(f(x))$is not defined. Being a composition function, the composition is always only defined in the domain of outer function. Try manipulating the data to find the upper and lower limits using given data.

Complete step by step solution:
Let us first of all note down all the given details and what exactly we need to find;
They have provided us with two function definitions along with their intervals also the values of constants $ \alpha $ and $ \beta $ .
The first function $ f(x) $ is defined as follows:
 $
  f(x) = \beta {x^2} + 3,\; - \infty < x < - 1 \\
  f(x) = 2x + \alpha ,\; - 1 \leqslant x < \infty \\
  $
The first function $ g(x) $ is defined as follows:
 $
  g(x) = x + 4,\;0 \leqslant x \leqslant 8 \\
  g(x) = - 3x - 2,\; - \infty < x < 0 \;
  $
Where the values of $ \alpha = 2,\;\beta = 3 $ .
We must find: the interval of the constants where the composition $ g(f(x)) $ is not defined.
Clearly for any composition function, we know that it is defined only in the domain of the outer function, here $ g(f(x)) $ is defined in the domain of $ g(x) $ . So we can write in this way;
 $ \Rightarrow 0 \leqslant f(x) \leqslant 4 $ and $ - 2 < f(x) < 0 $
Combining the two inequalities we can write;
 $ \Rightarrow - 2 < f(x) \leqslant 4 $
So now comparing with the domain of $ f(x) $ we can write the function $ f(x) $ is equal to;
 $ \Rightarrow f(x) = 2x + \alpha $ ,
but we know that the lower limit or minimum value $ x $ will take in that interval is $ - 1 $ , so $ f(x) $ becomes;
 $ \Rightarrow f( - 1) = \alpha - 2 $
And then we can write from the information given above, where the function is ‘not defined’ in is when;
 $ \Rightarrow a - 2 > 4 $
 $ \Rightarrow a > 6 \to (1) $
Also going back to definition of $ f(x) $ we know that when $ x < - 1 $ , function becomes;
 $ \Rightarrow f(x) = \beta {x^2} + 3 $
But again, clearly if $ \beta $ takes any value above $ 1 $ , the function exceeds the upper limit $ 4 $
So we can say that the function is not defined in the following region;
 $ \Rightarrow \beta > 1 \to (2) $
Therefore from equations $ (1),(2) $ , when combined we get;
The composition function is not defined when;
 $ \alpha \in (10,\infty ),\;\beta \in (5,\infty ) $
So the final right answer where the function is not defined is: option (i)
So, the correct answer is “Option (i)”.

Note: Let us remember that composition functions are defined as follows:
If we have any two functions $ h(x),j(x) $ , then its composition means;
 $ \Rightarrow h \circ j = h(j(x)) $
They satisfy certain properties:
* Commutative property: it means that the order of taking the composition does not matter, more clearly: $ \Rightarrow h \circ j = j \circ h $
* Associative property: if we have three functions, $ h(x),j(x),t(x) $ then associative property is as follows: $ \Rightarrow h \circ (j \circ t) = (h \circ j) \circ t $
* If a one-one function is taken then its composition function is also one-one.
* If an onto function is taken then its composition function is also always onto.
* $ {(h \circ j)^{ - 1}} = {h^{ - 1}} \circ {j^{ - 1}} $